I have one confusion when in question we knw tht speed is getting reduced so should we put the sign of acceleration negative?? For ex in this question:-

A car moving in a straight line at 30 m/s slows uniformly to a speed of 10 m/s in 5 sec. Determine:

(a)the acceleration of the car,

(b)displacement in the third second. When I take the sign of “a” negative the magnitude of a is coming out be positive and when I put the value of “a”(+ve) in b part i get wrong answer of b part

It's great that you're diving into the concepts of acceleration and displacement! Let's clarify how to approach this problem step by step, especially regarding the sign of acceleration when speed decreases.

Understanding Acceleration and Its Sign

Acceleration is defined as the rate of change of velocity over time. When an object slows down, we refer to this as negative acceleration or deceleration. This is why we assign a negative sign to acceleration when the speed decreases. In your example, since the car is slowing down from 30 m/s to 10 m/s, the acceleration will indeed be negative.

Calculating Acceleration

To find the acceleration, we can use the formula:

• a = (v_f - v_i) / t

Where:

• v_f = final velocity (10 m/s)
• v_i = initial velocity (30 m/s)
• t = time (5 seconds)

Now, substituting the values:

• a = (10 m/s - 30 m/s) / 5 s
• a = (-20 m/s) / 5 s
• a = -4 m/s²

So, the acceleration of the car is -4 m/s², indicating it is decelerating.

Finding Displacement in the Third Second

To determine the displacement during the third second, we can use the formula for displacement in uniformly accelerated motion:

• s = v_i * t + 0.5 * a * t²

However, since we want the displacement specifically during the third second, we can calculate the displacement for the first three seconds and then subtract the displacement for the first two seconds from it.

First, let's find the total displacement after 3 seconds:

• s(3) = v_i * 3 + 0.5 * a * (3)²
• s(3) = 30 m/s * 3 + 0.5 * (-4 m/s²) * (3)²
• s(3) = 90 m - 18 m
• s(3) = 72 m

Next, calculate the displacement after 2 seconds:

• s(2) = v_i * 2 + 0.5 * a * (2)²
• s(2) = 30 m/s * 2 + 0.5 * (-4 m/s²) * (2)²
• s(2) = 60 m - 8 m
• s(2) = 52 m

Now, to find the displacement during the third second:

• Displacement during the third second = s(3) - s(2)
• Displacement during the third second = 72 m - 52 m
• Displacement during the third second = 20 m

Final Thoughts

In summary, when the speed of an object decreases, the acceleration is negative. This negative value is crucial for correctly calculating displacement and other related quantities. In your case, using -4 m/s² for acceleration will lead you to the correct displacement during the third second, which is 20 m. If you have any more questions or need further clarification, feel free to ask!